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Physics · PYQ NEET UG 2016(1)
हिन्दीThe intensity at the maximum in a Young's double slit experiment is \( I_0 \). Distance between two slits is \( d = 5\lambda \), where \( \lambda \) is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance, \( D = 10d \)?
Correct answer · Option 3
\( \frac{1}{2} I_0 \)
Explanation
Path difference \( \Delta x = \frac{yd}{D} \). For \( y = d/2 \), \( \Delta x = \frac{d^2}{2D} = \frac{(5\lambda)^2}{2(50\lambda)} = \frac{25\lambda^2}{100\lambda} = \frac{\lambda}{4} \). Phase difference \( \phi = \frac{2\pi}{\lambda} \Delta x = \frac{\pi}{2} \). Intensity \( I = I_0 \cos^2(\phi/2) = I_0 \cos^2(\pi/4) = \frac{I_0}{2} \).
